Normal is perpendicular to tangent at the pint on the curve.
The slope of tangent to curve #y=f(x)# at #x=x+# is given by #f'(x_0)# and hence slope of normal is #-1/(f'(x_0))#. Further itpasses through point #(x_0,f(x_0))# on the curve. This gives us point slope form of normal to the curve as #y-f(x_0)=-1/(f'(x_0))(x-x_0)#.
Here #y=f(x)=sinx-e^xcotx# and #x_0=pi/3#
Hence #f(pi/3)=sin(pi/3)-e^(pi/3)cot(pi/3)=sqrt3/2-e^(pi/3)/sqrt3#
Further #f'(x)=cosx+e^xcsc^2x-e^xcotx#
and #f'(pi/3)=cos(pi/3)+e^(pi/3)csc^2(pi/3)-e^(pi/3)cot(pi/3)#
= #1/2+e^(pi/3)[4/3-1/sqrt3]=1/2+(4-sqrt3)/3e^(pi/3)#
and hence equation of normal is
#y-sqrt3/2+e^(pi/3)/sqrt3=-1/(1/2+(4-sqrt3)/3e^(pi/3))(x-pi/3)#
or #(y-sqrt3/2+e^(pi/3)/sqrt3)(1/2+(4-sqrt3)/3e^(pi/3))=pi/3-x#
